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Lagrange's theorem states that for any finite group $G$, the order (number of elements) of every subgroup $H$ of $G$ divides the order of $G$ .......(1)

The converse of Lagrange's theorem is if $x$ divides order of $G$ ,then there exists a sub group of order $x$. ...... (2)

If my statement is $p\to q$ then converse is $q\to p$.

i couldn't understand how converse of Lagrange's theorem comes...please explain with this definition of converse : $p\to q$ then converse is $q\to p$

Cloud JR K
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2 Answers2

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Actually, you need to write the Lagrange theorem in $p \rightarrow q$ form. So, the Lagrange theorem is actually,

If $H$ is a subgroup of $G$, then order of $H$ divides order of $G$.

Here $p:$ $H$ is a subgroup of $G$

$q;$ Order of $H$ divides order of $G$.

Here, actually the existence of $H$ is in the hypothesis ($p$) and the number $m$ is actually the order of the subgroup which divides order of $G$.

So, the converse will be

If a number $m$ divides order of $G$, then there is a subgroup of order $m$.

I hope this gives you insights.

Aniruddha Deshmukh
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  • It seems that the OP gets trouble in understanding a logical statement. – Mikasa Aug 12 '18 at 09:07
  • @ResidentDementor yes you are correct – Cloud JR K Aug 12 '18 at 09:08
  • Yes. Actually he does! Which is why I think that he should have made the tag of "logic" instead of "abstract algebra". – Aniruddha Deshmukh Aug 12 '18 at 09:08
  • well i change the tags.. thanks for your suggestion – Cloud JR K Aug 12 '18 at 09:10
  • So Lagrange's theorem can be restated as If there exists a subgroup H of order m ,then m | order of G.. Is this correct? – Cloud JR K Aug 12 '18 at 09:13
  • Absolutely correct! – Aniruddha Deshmukh Aug 12 '18 at 09:13
  • I'd write the converse of Lagrange's theorem like this: "For every finite group $G$, if $d$ divides $|G|$ then there exists a subgroup $H$ of $G$ with $|H|=d$ ". And so, to disprove this statement you just have to prove its negation which is, "There exists a finite group $G$ such that $d$ divides $|G|$ and there do not exist any subgroup $H$ of $G$ with $|H|=d$ ". To add further, one such finite group $G$ is $A_4$. – Aman Kushwaha Oct 20 '21 at 17:30
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Take the group $A_4$, then group of all even permutations of $4$ elements. $o(A_4)$ is $12$, but it doesn't have any subgroup of order $6$. Please check it !!

Anik Bhowmick
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